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My kids have this 3x4 cube puzzle, you know, the one where a picture is formed if you assemble the cubes correctly.

In reality you can create 6 different pictures, using different sides of the cubes.

How many possible ways are there to combine these cubes in a 3x4 pattern?

Or in other words, what is the probability to randomly put the cubes together and get a correct picture?

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  • $\begingroup$ So you have $12$ small (perfect) cubes that you can combine together into a single large (non-perfect) cube? $\endgroup$
    – barak manos
    Commented Jun 30, 2014 at 15:43
  • $\begingroup$ I think they form a $3\times 4 \times 1$ rectangle, and the picture is on the top face. $\endgroup$
    – vadim123
    Commented Jun 30, 2014 at 15:44
  • $\begingroup$ Not cube! They are lying flat on a 3x4 grid (landscape style, not that it matters :) $\endgroup$
    – stratosgear
    Commented Jun 30, 2014 at 15:44
  • $\begingroup$ as a matter of fact, that's almost exactly like the one I was talking about: i01.i.aliimg.com/photo/v1/1789407963/… :) $\endgroup$
    – stratosgear
    Commented Jun 30, 2014 at 15:48

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Consider first just the problem of orienting the correct faces upward. There are $6^{12}$ choices to make, of which only six are valid; doing this at random will give only a probability of $\frac{6}{6^{12}}\approx \frac{1}{360,000,000}$. Now we must arrange the twelve puzzle pieces into a $3\times 4$ grid correctly. There are $12!$ ways to do this, and only one is correct (up to rotation, which I will consider at the end). Then, after putting the twelve pieces into the correct locations, each piece must be oriented correctly, out of four possible ways. There are $4^{12}$ ways to do this, and again only one of these is correct. Putting this all together we get $$\frac{6}{6^{12}}\frac{1}{12!}\frac{1}{4^{12}}\approx 3.4\times 10^{-25}$$

However if they assemble the puzzle rotated $180^\circ$ we should count that as a solution too, so the answer really is twice that, or $6.9\times 10^{-25}$. For comparison, there are half a million times more positions of this puzzle than there are for Rubik's cube, which has a paltry $4\times 10^{-19}$ probability of a randomly chosen position being solved.

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  • $\begingroup$ How about the fact that there are 6 possible images that can be constructed (using different sides of the cubes?) Doesn't it matter? $\endgroup$
    – stratosgear
    Commented Jun 30, 2014 at 15:53
  • $\begingroup$ And assuming that a 180 degree solution is a valid solution, shouldn't we divide by 2, instead of multiplying by two..? $\endgroup$
    – stratosgear
    Commented Jun 30, 2014 at 15:54
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    $\begingroup$ 1. six images was accounted for in the numerator of $\frac{6}{6^{12}}$; otherwise it would be $\frac{1}{6^{12}}$. 2. We multiply by two because the result is the probability of getting an answer "at random", which is doubled since there are two answers that work. $\endgroup$
    – vadim123
    Commented Jun 30, 2014 at 15:57
  • $\begingroup$ Great, thanks, make sense. So they have a much better probability to solve the Rubik's cube than that Disney puzzle. How deceiving!!! $\endgroup$
    – stratosgear
    Commented Jun 30, 2014 at 16:00
  • $\begingroup$ Well, yes and no. The hard part of Rubik is that you can't just arrange the cubes how you want, there are only certain moves allowed. However if you tried to solve both puzzles blindly by random moves, then yes the Rubik's cube is easier to solve. $\endgroup$
    – vadim123
    Commented Jun 30, 2014 at 16:01

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